#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Software: PyCharm
# @Version : Python-
# @Author  : Shengji He
# @Email   : hsjbit@163.com
# @File    : LongestMountainArray.py
# @Time    : 2020/10/25 15:27
# @Description:
from typing import List


class Solution:
    def longestMountain(self, A: List[int]) -> int:
        """
        Let's call any (contiguous) subarray B (of A) a mountain if the following properties hold:
            - B.length >= 3
            - There exists some
            0 < i < B.length - 1 such that B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]
        (Note that B could be any subarray of A, including the entire array A.)
        
        Given an array A of integers, return the length of the longest mountain. 
        
        Return 0 if there is no mountain.
        
        Example 1:
            Input: [2,1,4,7,3,2,5]
            Output: 5
            Explanation: The largest mountain is [1,4,7,3,2] which has length 5.
        Example 2:
            Input: [2,2,2]
            Output: 0
            Explanation: There is no mountain.
        Note:
            1. 0 <= A.length <= 10000
            2. 0 <= A[i] <= 10000
        Follow up:
            - Can you solve it using only one pass?
            - Can you solve it in O(1) space?

        :param A:
        :return:
        """
        n = len(A)
        ans = left = 0
        while left + 2 < n:
            right = left + 1
            if A[left] < A[left + 1]:
                while right + 1 < n and A[right] < A[right + 1]:
                    right += 1
                if right < n - 1 and A[right] > A[right + 1]:
                    while right + 1 < n and A[right] > A[right + 1]:
                        right += 1
                    ans = max(ans, right - left + 1)
                else:
                    right += 1
            left = right
        return ans


if __name__ == '__main__':
    S = Solution()
    A = [2, 1, 4, 7, 3, 2, 5]
    print(S.longestMountain(A))
    print('done')
